I am looking to prove the following identity combinatorially:
$\sum_k$ $n \choose 2k$ $2k \choose k$ $2^{n-2k}$ = $2n \choose n$
Clearly the RHS counts the number of ways to choose n elements out of a set containing an even number of elements. Specifically, we choose half the set. The LHS counts the same by instead choosing all possible even subsets of n (those with even cardinality) and then choosing whether the remaining terms are in or out of this subset. However, I am not sure how the second term fits in then.